I am assuming that you are trying to prove that the sum has at least 3 PRIME factors (including multiplicities), and not just at least 3 factors, because since the sum of 2 odd numbers is even, the sum will obviously have at least 3 factors (1, 2 and the number itself will all be factors) So let's assume that you wish to prove the deeper result, which is that the sum will have (including multiple copies of the same prime factor) at least 3 PRIME factors. Let's start with an example: 7 and 11 are consecutive primes, and 7+11=18, and the prime factors of 18 are 2, 3, 3. Ok, so far so good. Let's try another example: 41 and 43 are consecutive primes, and 41+43 =84, and the prime factors of 84 are 2,2,3,7 which is 4 prime factors, which is good. Ok, so now how do we prove this?
Let A and B be consecutive odd primes, with A<B. So this means that if a number is inbetween A and B, it cannot be prime. Since A and B are both odd, A+B must be even, so A+B will be divisible by 2 so A+B=2*X for some number X (meaning that 2 is a prime factor of A+B). Note that this means that X is the average of A and B, so A<X<B (since the average of two numbers is between them). Therefore, X cannot be prime, because there are no primes between A and B (A and B are CONSECUTIVE primes) Thus X can itself be written as a product of at least 2 primes, and so A+B will have at least 3 prime factors: the 2 from the fact that A+B is even, and the at lest two prime factors of X. Done.